# Solving Leetcode Interviews in Seconds with AI: Maximum Value after Insertion


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1881" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given a very large integer n, represented as a string,​​​​​​ and an integer digit x. The digits in n and the digit x are in the inclusive range [1, 9], and n may represent a negative number. You want to maximize n's numerical value by inserting x anywhere in the decimal representation of n​​​​​​. You cannot insert x to the left of the negative sign.  For example, if n = 73 and x = 6, it would be best to insert it between 7 and 3, making n = 763. If n = -55 and x = 2, it would be best to insert it before the first 5, making n = -255.  Return a string representing the maximum value of n​​​​​​ after the insertion.   Example 1:  Input: n = "99", x = 9 Output: "999" Explanation: The result is the same regardless of where you insert 9.  Example 2:  Input: n = "-13", x = 2 Output: "-123" Explanation: You can make n one of {-213, -123, -132}, and the largest of those three is -123.    Constraints:  1 <= n.length <= 105 1 <= x <= 9 The digits in n​​​ are in the range [1, 9]. n is a valid representation of an integer. In the case of a negative n,​​​​​​ it will begin with '-'.  

	# Explanation
	Here's the approach, complexity analysis, and Python code:

*   **Handle the sign:** Determine if the number is positive or negative.
*   **Iterate and compare:** Iterate through the digits of the number (excluding the negative sign, if present). Insert `x` at the first position where it leads to a larger number. For positive numbers, insert `x` if a digit is smaller than `x`. For negative numbers, insert `x` if a digit is *larger* than `x`.
*   **Append at the end (if necessary):** If no suitable insertion point is found during iteration, append `x` to the end of the number.

*   **Time Complexity:** O(n), where n is the length of the string `n`.
*   **Space Complexity:** O(n), to store the potentially modified string.

	
	# Code
	```python
	def maxValue(n: str, x: int) -> str:
    sign = 1 if n[0] != '-' else -1
    result = ""
    inserted = False

    if sign == 1:
        for i in range(len(n)):
            if int(n[i]) < x:
                result += str(x) + n[i:]
                inserted = True
                break
            else:
                result += n[i]
    else:
        result += "-"
        for i in range(1, len(n)):
            if int(n[i]) > x:
                result += str(x) + n[i:]
                inserted = True
                break
            else:
                result += n[i]

    if not inserted:
        result += str(x)

    return result
	```